Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Mechanical Properties of Solids for CBSE Class 11 Physics.
Key Concepts
Elasticity and Solids
How materials respond to forces
1
Stress
Formula
The internal restoring force set up per unit area of a deformed body. It opposes the deforming force. SI unit is $\text{N/m}^2$ or Pascal ($\text{Pa}$).
$$\sigma = \frac{F}{A}$$
2
Strain
Formula
The ratio of the change in configuration (length, volume, or shape) to the original configuration. It is a dimensionless and unitless quantity.
$$\text{Longitudinal Strain } \epsilon = \frac{\Delta L}{L} \quad | \quad \text{Volume Strain} = \frac{\Delta V}{V}$$
3
Hooke’s Law
Formula
For small deformations (within the proportional limit), the stress induced in a body is directly proportional to the strain produced.
$$\text{Stress} \propto \text{Strain} \implies \text{Stress} = E \times \text{Strain}$$
4
Young’s Modulus ($Y$)
Formula
The ratio of longitudinal stress to longitudinal strain. It characterizes the stiffness of a solid material against length changes.
$$Y = \frac{F/A}{\Delta L/L} = \frac{FL}{A \Delta L}$$
5
Bulk Modulus ($B$) & Compressibility ($k$)
Formula
Bulk modulus is the ratio of hydraulic stress (pressure) to volume strain. The negative sign shows volume decreases as pressure increases. Compressibility is its reciprocal.
$$B = -\frac{P}{\Delta V/V} \quad | \quad k = \frac{1}{B}$$
6
Shear Modulus / Modulus of Rigidity ($G$)
Formula
The ratio of tangential (shearing) stress to the corresponding shearing strain ($\theta$). It measures resistance to shape change.
$$G = \frac{F/A}{\theta} \quad \text{where } \theta \approx \frac{\Delta x}{L}$$
7
Poisson’s Ratio ($\sigma$)
Formula
When a wire is stretched, it becomes thinner. Poisson’s ratio is the ratio of lateral (transverse) strain to longitudinal strain.
$$\sigma = \frac{\Delta d / d}{\Delta L / L}$$
8
Elastic Potential Energy
Formula
The work done in stretching a wire is stored as elastic potential energy ($U$). Energy density ($u$) is the energy stored per unit volume.
$$U = \frac{1}{2} F \Delta L = \frac{1}{2} (\text{Stress}) \times (\text{Strain}) \times \text{Volume}$$
$$u = \frac{1}{2} \text{Stress} \times \text{Strain} = \frac{1}{2} Y (\text{Strain})^2$$
Concept Deep Dive
01
Reading the Stress-Strain Curve
The fingerprint of a materialIf you graph Stress (Y-axis) vs. Strain (X-axis) while stretching a metal wire, it reveals the material’s structural journey:
1. Proportional Region (O to A): The graph is a straight line. Hooke’s Law is obeyed. The slope of this line equals Young’s Modulus ($Y$).
2. Elastic Limit / Yield Point (B): The maximum stress the material can take and still return to its original length when released. Past this point, it suffers permanent deformation (plasticity).
3. Ultimate Tensile Strength (D): The absolute maximum stress the material can withstand before “necking” (thinning out) begins.
4. Fracture Point (E): The wire completely snaps and breaks.
1. Proportional Region (O to A): The graph is a straight line. Hooke’s Law is obeyed. The slope of this line equals Young’s Modulus ($Y$).
2. Elastic Limit / Yield Point (B): The maximum stress the material can take and still return to its original length when released. Past this point, it suffers permanent deformation (plasticity).
3. Ultimate Tensile Strength (D): The absolute maximum stress the material can withstand before “necking” (thinning out) begins.
4. Fracture Point (E): The wire completely snaps and breaks.
Ductile vs. Brittle
If the gap between the Yield Point and Fracture Point is very large, the material is Ductile (can be drawn into long wires like Copper). If the gap is tiny and it breaks almost immediately after the yield point, it is Brittle (like Glass or Cast Iron).
02
Why is Steel More Elastic than Rubber?
Redefining “Elasticity”In everyday language, “elastic” means easily stretchy. In physics, Elasticity is the resistance to change, and the ability to strongly snap back.
If you hang a $10 \text{ kg}$ weight from a steel wire and a rubber cord of the same thickness, the rubber stretches a lot (high strain), while the steel barely stretches at all (low strain). Because Young’s Modulus $Y = \frac{\text{Stress}}{\text{Strain}}$, a lower strain means a much higher $Y$. Steel opposes the deformation much more aggressively than rubber. Therefore, in physics, Steel is highly elastic, and Rubber is far less elastic!
If you hang a $10 \text{ kg}$ weight from a steel wire and a rubber cord of the same thickness, the rubber stretches a lot (high strain), while the steel barely stretches at all (low strain). Because Young’s Modulus $Y = \frac{\text{Stress}}{\text{Strain}}$, a lower strain means a much higher $Y$. Steel opposes the deformation much more aggressively than rubber. Therefore, in physics, Steel is highly elastic, and Rubber is far less elastic!
Compare & Contrast
✗ Elasticity
- The property to regain original shape and size after the deforming force is removed.
- Occurs before the Yield Point.
- Involves temporary stretching of atomic bonds without breaking them.
- Example: Spring, Quartz, Steel.
✓ Plasticity
- The inability to regain original shape; permanent deformation occurs.
- Occurs after the Yield Point.
- Involves atoms slipping past each other into new permanent positions.
- Example: Putty, Clay, Dough.
Remember
Elastomers (like the rubber in a balloon or the aorta in your heart) are a special third category. They can stretch to many times their original length and return, but they do not obey Hooke’s Law (their stress-strain graph is not a straight line).
Common Mistakes to Avoid
Mistake 1
Confusing Stress with Pressure: Both have the formula $F/A$ and the unit Pascals ($\text{Pa}$). However, Pressure is an external force applied normal to a surface. Stress is an internal restoring force developed within the body. Also, pressure is always compressive (pushing in), while stress can be tensile (pulling out) or shearing.
Mistake 2
Using Final Length instead of Original Length: In the formula $\epsilon = \Delta L / L$, the denominator $L$ is strictly the original, unstretched length. If a $2 \text{ m}$ wire stretches to $2.05 \text{ m}$, the strain is $0.05 / 2.0$, NOT $0.05 / 2.05$.
Mistake 3
Thinking Breaking Stress depends on thickness: “Breaking Force” depends on the thickness (Area) of a wire (a thicker wire takes more force to snap). But “Breaking Stress” ($F_{\text{break}}/A$) is a constant property of the material itself. It does not change whether the wire is thick, thin, long, or short.
Exam Tips
Tip 1
Thermal Stress Shortcut: If a metal rod of length $L$, Young’s modulus $Y$, and coefficient of linear expansion $\alpha$ is clamped rigidly at both ends and heated by $\Delta T$, it wants to expand but cannot. The thermal strain is $\alpha \Delta T$, and the thermal stress developed is simply $\sigma = Y \alpha \Delta T$.
Tip 2
When calculating the elongation of a heavy wire due to its own weight (hanging vertically), assume the entire weight acts at its center of gravity (length $L/2$). The formula becomes $\Delta L = \frac{MgL}{2AY} = \frac{\rho g L^2}{2Y}$ where $\rho$ is density.
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Mechanical Properties · CBSE Exam1
Why do bridge pillars have a cross-section of an I-shape (I-beams)? [1 mark]
Answer
To minimize depression (bending) while saving material.
📝
Explanation
The depression ($\delta$) of a beam loaded at the center is given by $\delta = \frac{W L^3}{4 b d^3 Y}$, where $b$ is breadth and $d$ is depth (thickness). To minimize bending, depth ($d$) is much more effective than breadth ($b$) because it is cubed. An I-beam provides a large depth ($d$) and concentrates the mass at the top and bottom flanges where stress is maximum, offering high strength while reducing the beam’s own dead weight and saving steel.
2
A structural steel rod has a radius of $10 \text{ mm}$ and a length of $1.0 \text{ m}$. A $100 \text{ kN}$ force stretches it along its length. Calculate the (a) stress, and (b) elongation. (Young’s modulus of structural steel $Y = 2.0 \times 10^{11} \text{ N/m}^2$) [3 marks]
Answer
(a) $3.18 \times 10^8 \text{ Pa}$ (b) $1.59 \text{ mm}$
📝
Explanation
Given: $r = 10 \text{ mm} = 10 \times 10^{-3} \text{ m} = 0.01 \text{ m}$. $F = 100 \text{ kN} = 10^5 \text{ N}$. $L = 1 \text{ m}$.
(a) Stress: $\sigma = \frac{F}{A} = \frac{F}{\pi r^2}$
$\sigma = \frac{10^5}{3.14 \times (0.01)^2} = \frac{10^5}{3.14 \times 10^{-4}} \approx 3.18 \times 10^8 \text{ N/m}^2$ (or Pa).
(b) Elongation: From $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\sigma}{\Delta L / L}$
$\Delta L = \frac{\sigma \times L}{Y} = \frac{3.18 \times 10^8 \times 1}{2.0 \times 10^{11}} = 1.59 \times 10^{-3} \text{ m} = 1.59 \text{ mm}$.
3
Compute the fractional change in volume of a glass slab when subjected to a hydraulic pressure of $10 \text{ atm}$. (Bulk modulus of glass $B = 37 \times 10^9 \text{ N/m}^2$, $1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}$). [2 marks]
Answer
$2.73 \times 10^{-5}$
📝
Explanation
Given: $P = 10 \text{ atm} = 10 \times 1.013 \times 10^5 = 1.013 \times 10^6 \text{ Pa}$.
Bulk Modulus $B = 37 \times 10^9 \text{ Pa}$.
The fractional change in volume is the Volume Strain ($\Delta V/V$).
Using the formula $B = \frac{P}{\Delta V/V}$ (ignoring negative sign for magnitude):
$\frac{\Delta V}{V} = \frac{P}{B} = \frac{1.013 \times 10^6}{37 \times 10^9} \approx 0.0273 \times 10^{-3} = 2.73 \times 10^{-5}$.
(This shows glass is highly incompressible!).
Concept Map
Mechanical Properties of Solids connects to →
Properties of Matter
Fluid Mechanics (Bulk Modulus)
Oscillations (Springs & SHM)
Work & Energy (Stored Potential)
Thermodynamics (Thermal expansion)
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